2.4 Annihilators, Complemented Subspaces

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1 40 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY 2.4 Annihilators, Complemented Subspaces Definition (Annihilators, Pre-Annihilators) Assume X is a Banach space. Let M X and N X. We call the annihilator of M and the pre-annihilator of N. M? = {x 2 X : 8x2M hx,xi =0} X, N? = {x 2 X : 8x 2N hx,xi =0} X, Proposition Let X be a Banach space, and assume M X and N X. a) M? is a closed subspace of X, M? =(span(m))?,and(m? )? = span(m), b) N? is a closed subspace of X, N? = (span(n))?,andspan(n) (N? )?. c) span(m) =X () M? = {0}. Proposition If X is Banach space and Y X is a closed subspace then (X/Y ) is isometrically isomorphic to Y? via the operator :(X/Y )! Y?, with (z )(x) =z (x). (recall x := x + Y 2 X/Y for x 2 X). Proof. Let Q : X! X/Y be the quotient map. For z 2 (X/Y ), (z ), as defined above, can be written as (z )= z Q. Thus (z ) 2 X. Since Q(Y )={0} it follows that (z ) 2 Y?. For z 2 (X/Y ) we have k (z )k = sup hz,q(x)i = sup hz, xi = kz k (X/Y ), x2b X x2b X/Y where the second equality follows on the one hand from the fact that kq(x)k apple kxk, for x 2 X, and on the other hand, from the fact that for any x = x+y 2 X/Y there is a sequence (y n ) Y so that lim sup n!1 kx + y n k = k xk. Thus is an isometric embedding. If x 2 Y? X,wedefine z : X/Y! K, x+ Y 7! hx,xi.

2 2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES 41 First note that this map is well defined (since hx,x+ y 1 i = hx,x+ y 2 i for y 1,y 2 2 Y ). Since x is linear, z is also linear, and hz, xi = hx,xi, for all x 2 X, and thus kz k (X/Y ) = kx k. Finally, since h (z ),xi = hz,q(x)i = hx,xi, it follows that (z )=x, and thus that is surjective. Proposition Assume X and Y are Banach spaces and T 2 L(X, Y ). Then (2.2) (2.3) T (X)? = N (T ) and T (Y ) N(T )? T (X) =N (T )? and T (Y )? = N (T ). Proof. We only prove (2.2). The verification of (2.3) is similar. For y 2 Y y 2T (X)? () 8x2X hy,t(x)i =0 () 8x2X ht (y ),xi =0 () T (y )=0 () y 2N(T ), which proves the first part of (2.2), and for y 2 Y and all x 2N(T ), it follows that ht (y ),xi = hy,t(x)i = 0, which implies that T (Y ) N (T )?, and, thus, T (X ) N(T ). Definition Let X be a Banach space and let U and V be two closed subspaces of X. We say that X is the complemented sum of U and V and we write X = U V, if for every x 2 X there are u 2 U and v 2 V, so that x = u + v and so that this representation of x as sum of an element of U and an element of V is unique. We say that a closed subspace Y of X is complemented in X if there is a closed subspace Z of X so that X = Y Z. Remark. Assume that the Banach space X is the complemented sum of the two closed subspaces U and V. We note that this implies that U \ V = {0}. We can define two maps P : X! U and Q : X! V where we define P (x) 2 U and Q(x) 2 V by the equation x = P (x)+q(y), with P (x) 2 U and Q(x) 2 V (which, by assumption, has a unique solution). Note that P and Q are linear. Indeed if P (x 1 )=u 1, P (x 2 )=u 2, Q(x 1 )=v 1, Q(x 2 )=v 2, then for, µ2k we have x 1 +µx 2 = u 1 +µu 2 + v 1 +µv 2, and thus, by uniqueness P ( x 1 +µx 2 )= u 1 +µu 2, and Q( x 1 +µx 2 )= v 1 +µv 2. Secondly it follows that P P = P, and Q Q = Q. Indeed, for any x 2 X we we write P (x) =P (x)+02 U + V, and since this representation

3 42 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY of P (x) is unique it follows that P (P (x)) = P (x). The argument for Q is the same. Finally it follows that, again using the uniqueness argument, that P is the identity on U and Q is the identity on V. We therefore proved that a) P is linear, b) the image of P is U c) P is idempotent, i.e. P 2 = P We say in that case that P is a linear projection onto U. Similarly Q is a a linear projection onto V, and P and Q are complementary to each other, meaning that P (X)\Q(X) ={0} and P +Q = Id. A linear map P : X! X with the properties (a) and (c) is called projection. The next Proposition will show that P and Q as defined in above remark are actually bounded. Lemma Assume that X is the complemented sum of two closed subspaces U and V. Then the projections P and Q as defined in above remark are bounded. Proof. Consider the norm on X defined by x = kp (x)k + kq(x)k, for x 2 X. We claim that (X, ) is also a Banach space. Indeed if (x n ) X with x n = kp (x n )k + kq(x n )k < 1. Then u = P 1 P (x n) 2 U, v = P 1 Q(x n) 2 V (U and V are assumed to be closed) converge in U and V (with respect to k k, respectively, and since k kapple also x = P 1 x n converges and x = x n = lim n!1 P (x j )+Q(x j )= lim n!1 P (x j )+ lim n!1 Q(x j )=u+v, and x x n = u P (x n )+v Q(x n )

4 2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES 43 = u P (x n ) + v Q(x n )! n!1 0, (here all series are meant to converge with respect to k k) which proves that (X, ) is complete. Since the identity is a bijective linear bounded operator from (X, )to (X, k k) it has by Corollary of the Closed Graph Theorem a continuous inverse and is thus an isomorphy. Since kp (x)k apple x and kq(x)k apple x we deduce our claim. Proposition Assume that X is a Banach space and that P : X! X, is a bounded projection onto a closed subspace of X. Then X = P (X) N(P ). Theorem There is no linear bounded operator T : `1! `1 so that the kernel of T equals to c 0. Corollary c 0 is not complemented in `1. Proof of Theorem For n 2 N we let e n be the n-th coordinate functional on `1, i.e. e n : `1! K, x =(x j ) 7! x n. Step 1. If T : `1! `1 is bounded and linear, then Indeed, note that N (T )= 1\ N (e n T ). x 2N(T ) () 8n2N e n(t (x)) = he n,t(x)i =0. In order to prove our claim we will show that c 0 cannot be the intersection of the kernel of countably many functionals in ` 1. Step 2. There is an uncountable family (N : 2 I) of infinite subsets of N for which N \ N is finite whenever 6= are in I. Write the rational numbers Q as a sequence (q j : j 2 N), and choose for each r 2 R asequence(n k (r) :k 2N), so that (q nk (r) : k 2N) converges to r. Then, for r 2R let N r = {n k (r) :k 2N}. For i2i, putx =1 N 2`1, i.e. x =( ( ) k : k 2N) with ( ) k = ( 1 if k 2 N 0 if k 62 N.

5 44 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY Step 3. If f 2 ` 1 and c 0 N(f) then{ 2 I : f(x ) 6= 0} is countable. In order to verify Step 3 let A n = { : f(x ) 1/n}, for n 2 N. It is enough to show that for n 2 N the set A n is finite. To do so, let 1, 2,..., k be distinct elements of A n and put x = P k sign f(x j ) x j (for a 2 C we put sign(a) =a/ a ) and deduce that f(x) k/n. Now consider M j = N j \ S i6=j N i.thenn j \ M j is finite, and thus it follows for kx x = sign(f(x j ))1 Mj that f(x) =f( x) (sincex x 2 c 0 ). Since the M j, j =1, 2...k are pairwise disjoint, it follows that k xk 1 = 1, and thus k apple f(x) =f( x) applekfk. n Which implies that that A n can have at most nkfk elements. Step 4. If c 0 T 1 N (f n), for a sequence (f n ) ` 1, then there is an 2 I so that x 2 T 1 N (f n). In particular this implies that c 0 6= T n2n N (f n). Indeed, Step 3 yields that C = { 2 I : f n (x ) 6= 0 for some n 2 N} = [ n2n{ 2 I : f n (x ) 6= 0}, is countable, and thus I \ C is not empty. Remark. Assume that Z is any subspace of `1 which is isomorphic to c 0, then Z is not complemented. The proof of that is a bit harder. Theorem [So] Assume Y is a subspace of a separable Banach space X and T : Y! c 0 is linear and bounded. Then T can be extended to a linear and bounded operator T : X! c 0. Moreover, T can be chosen so that k T kapple2kt k. Corollary Assume that X is a separable Banach space which contains a subspace Y which is isomorphic to c 0. Then Y is complemented in X. Proof. Let T : Y! c 0 be an isomorphism. Then extend T to T : X! c 0 and put P = T 1 T.

6 2.4. ANNIHILATORS, COMPLEMENTED SUBSPACES 45 Proof of Theorem Note that an operator T : Y! c 0 is defined by a (Y,Y)nullsequence(y n) Y,i.e. T : Y! c 0, y 7! (hy n,yi : n 2 N). We would like to use the Hahn Banach theorem and extend each y n to an element x n 2 X n,withky nk = kx nk, and define T (x) :=(hx n,xi : n 2 N), x2x. But the problem is that (x n) might not be (X,X) convergent to 0, and thus we can only say that (hx n,xi : n 2 N) 2 `1, but not necessarily in c 0. Thus we will need to change the x n somehow so that they are still extensions of the yn but also (X,X)null. Let B = kt kb X. B is (X,X)-compact and metrizable (since X is separable). Denote the metric which generates the (X,X)-topology by d(, ). Put K = B \ Y?. Since Y? X is (X,X)-closed, K is compact and every (X,X)-accumulation point of (x n) lies in K. Indeed, this follows from the fact that x n(y) =yn(y)! n!1 0. This implies that lim n!1 d(x n,k) = 0, thus we can choose (zn) K so that lim n!1 d(x n,zn) = 0, and thus (x n zn)is (X,X)-null and for y 2 Y it follows that hx n zn,yi = hx n,yi, n2n. Choosing therefore yields our claim. T : X! c 0, x 7! (hx n z n,xi : n 2 N), Remark. Zippin [Zi] proved the converse of Theorem: if Z is an infinitedimensional separable Banach space admitting a projection from any separable Banach space X containing it, then Z is isomorphic to c 0. Exercises 1. Prove Proposition a) Assume that `1 isomorphic to a subspace Y of some Banach space X, then Y is complemented in X. b) Assume Z is a closed subspace of a Banach space X, and T : Z! `1 is linear and bounded. Then T can be extended to a linear and bounded operator T : X! `1, withk T k = kt k. 3. Show that for a Banach space X, the dual space X is isometrically isomorphic to complemented subspace of X, via the canonical embedding.

7 46 CHAPTER 2. WEAK TOPOLOGIES AND REFLEXIVITY 4. Prove Proposition Prove (2.3) in Proposition